Question:

\(3\times10^{22}\) molecules of \(Na_2CO_3\) molecular weight \(=106\) present in \(500\text{ ml}\) of solution. The normality of the solution formed is \(\left(N=6\times10^{23}\text{ mol}^{-1}\right)\)

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First convert molecules into moles using Avogadro number. Then find molarity and multiply by valency factor to get normality.
  • \(0.1N\)
  • \(0.2N\)
  • \(0.4N\)
  • \(0.05N\)
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The Correct Option is B

Solution and Explanation

Number of molecules of sodium carbonate is: \[ 3\times10^{22}. \] Avogadro number is: \[ 6\times10^{23}. \] Number of moles is: \[ \text{Moles}=\frac{3\times10^{22}}{6\times10^{23}}. \] \[ =\frac{3}{6}\times10^{-1}. \] \[ =0.5\times0.1. \] \[ =0.05\text{ mol}. \] Volume of solution is: \[ 500\text{ ml}=0.5\text{ L}. \] Molarity is: \[ M=\frac{\text{moles}}{\text{volume in litres}}. \] \[ M=\frac{0.05}{0.5}. \] \[ M=0.1M. \] For \(Na_2CO_3\), the valency factor is \(2\), because carbonate ion has charge \(2-\). Normality is: \[ N=M\times \text{valency factor}. \] \[ N=0.1\times2. \] \[ N=0.2N. \] Hence, the normality is: \[ 0.2N. \]
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